# What is the sum of the series $\sum \limits_{k=0}^\infty q^{2^k}$ if $|q|\lt1$

I wonder if anybody knows how to calculate the series below? $\sum \limits_{k=0}^\infty q^{2^k}$ if $|q|\lt1$?

Thanks a lot for answers.

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A bit of discussion is contained on the Wikipedia page on Lacunary series. – t.b. Jan 5 '12 at 7:27
@Mathlover: Notice that several users have tried to improve your post (mostly by adding LaTeX for proper rendering of math symbols). You should check if we did not unintentionally change the meaning of your question and, if necessary, edit your question again. – Martin Sleziak Jan 5 '12 at 7:27
you may want to see this paper: repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/62866/1/… – deoxygerbe Jan 5 '12 at 9:01

Thanks a lot for links

$\sum_{k=0}^\infty q^{2^k}$ if $|q|\lt1$

I also have noticed that the series above is related to the function equation

$G(2x)-G(x)=e^x-1$.

$$G(x)=\sum_{k=1}^\infty \frac{x^k}{k!(2^k-1)}$$

$\lim_{n\to\infty} (n+1+G(x.2^{n+1})-G(x)) = \ e^x+e^{2x}+e^{4x}+e^{8x}+\dots$ where $q=e^x$

And also other relation of that series,

$e^x+e^{2x}+e^{4x}+e^{8x}+\dots=F(x)$ where $q=e^x$

$x=2^t$

$e^{2^t}+e^{2^{t+1}}+e^{2^{t+2}}+\dots=F(2^t)=H(t)$

$e^{2^t}+H(t+1)=H(t)$

$H(t+1)-H(t)=-e^{2^t}$

$H(t)= k-\int_0^t e^{2^t} \mathrm{d}t+ \frac12 e^{2^t}-\frac1{12} \frac{\partial}{\partial t}(e^{2^t})+\dots$

Here need to find $k$ to complete the function $H(t)$ ( constants are bernoulli numbers)

Please advice me if anything wrong in my relations and need a hint to find $k$.

In other way, I found the way to find exact solution via Fourier transfom

$H(t+1)-H(t)=-e^{2^t}$ //We need to take fourier integral both side

$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t-\int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t=-\int_{-\infty}^{+\infty} e^{2^{t}}e^{-2πift} \mathrm{d}t$

$V(f)= \int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t$

$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t=\int_{-\infty}^{+\infty} H(z)e^{-2πif(z-1)} \mathrm{d}z=V(f).e^{2πif}$

$e^{2πif}V(f)-V(f)=-\int_{-\infty}^{+\infty} e^{2^{t}}e^{-2πift} \mathrm{d}t$

$V(f)=-\int_{-\infty}^{+\infty} (e^{2^{t}}e^{-2πift})/(e^{2πif}-1) \mathrm{d}t$

//Now We need to take inverse fourier transform

$H(z)=\int_{-\infty}^{+\infty} V(f) e^{2πifz} \mathrm{d}f=\int_{-\infty}^{+\infty} e^{2πifz}(-\int_{-\infty}^{+\infty} (e^{2^{t}}e^{-2πift})/(e^{2πif}-1) \mathrm{d}t).\mathrm{d}f$

$H(z)=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{2πifz} (e^{2^{t}}e^{-2πift})/(1-e^{2πif}) \mathrm{d}t\mathrm{d}f$
Exact solution of H(t) in integral representive way

$\sum_{k=0}^\infty q^{2^k}=H(t)=H(log_2(lnq))$

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What you posted here was very hard to read (at least for me). I've tried to add a new version (but I've kept the old one, since I might have made some mistake.) Please, check whether this is what you meant and do any changes, that are needed. If you click on edit and view the source, you will see that writing basic mathematics, such as integrals, sums, fractions, exponents, is not that difficult. When you are editing, you have a question mark there. By clicking on it, you can get to help sections, which mentions help on LaTeX, too. – Martin Sleziak Jan 6 '12 at 8:57
Thanks a lot Martin for your help – Mathlover Jan 6 '12 at 9:07